\newproblem{lay:1_7_9}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.7.9}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Given the vectors $\mathbf{v}_1=\begin{pmatrix}1\\-3\\2\end{pmatrix}$, $\mathbf{v}_2=\begin{pmatrix}-3\\9\\-6\end{pmatrix}$, and
	$\mathbf{v}_3=\begin{pmatrix}5\\-7\\h\end{pmatrix}$. For which value of $h$ is the set $S=\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}$ 
	linearly dependent.
}{
   % Solution
	We need to solve the vector equation
	\begin{center}
		$x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{0}$
	\end{center}
	and find a non-trivial solution. The augmented matrix of this equation system is
	\begin{center}
		$\left(\begin{array}{rrr|r}
		    1 & -3 &  5 & 0 \\
			 -3 &  9 & -7 & 0 \\
			  2 & -6 &  h & 0 \\
		\end{array}\right)\sim
		\left(\begin{array}{rrr|r}
		    1 & -3 &  5 & 0 \\
			  0 &  0 &  8 & 0 \\
			  0 &  0 &  0 & 0 \\
		\end{array}\right)$
	\end{center}
	This equation system is compatible indeterminate for any value of $h$, meaning that the set of vectors is linearly dependent
	disregarding the value of $h$. This is because $\mathbf{v}_2=-3\mathbf{v}_1$.
}
\useproblem{lay:1_7_9}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
